If $T:H \to H$ is linear and bounded and compact and $\{h_n\}$ is bounded, is $\{T(h_n)\}$ a compact subset of $H$? We have $H$ as a Hilbert space.
I am getting a problem with definition of compact operator since it maps a bounded set only into a relatively compact set
On
Remember that a finite-dimensional vector space with an inner product is a perfectly good Hilbert space, and all its linear operators are compact. So you can get a counterexample by taking $H$ to be $\mathbb R$ or $\mathbb C$ (depending on whether you prefer real or complex Hilbert spaces), taking $T$ to be the identity map, and taking $h_n=\frac1n$.
If, for some reason, you insist on an infinite-dimensional Hilbert space, then let $H$ be one (for example $l^2$), and let $T:H\to H$ be the projection to a one-dimensional subspace. That's compact, and then you can use my previous example of $h_n$ in that one-dimensional subspace.
Not necessarily. Take $H = \ell^2$ and let $(Th)(i) = \frac{1}{i} h(i)$. You can verify that $T$ is compact. Let $\{h_n\}$ be the usual orthonormal basis for $\ell^2$, i.e. $h_n(i) = \delta_{ni}$, so that $T h_n = \frac{1}{n} h_n$. Then $T h_n \to 0$, and so $\{T(h_n)\}$ is not closed.